\(n_{CuO}=4a\left(mol\right)\Rightarrow n_{FeO}=a\left(mol\right)\)
\(m_X=80\cdot4a+72a=19.6\left(g\right)\)
\(\Rightarrow a=0.05\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(FeO+H_2\underrightarrow{^{^{t^0}}}Fe+H_2O\)
\(m_{cr}=0.2\cdot64+0.05\cdot56=15.6\left(g\right)\)
\(V_{H_2}=\left(0.05\cdot4+0.05\right)\cdot22.4=5.6\left(l\right)\)